CODEWITHSUNDEEP
pythonpython-2.7python-3.x
COD3R
COD3R

Reputation: 129

In list how to count string in a sequence

I have a list with duplicate elements. I want to count the occurrence of each element

example-> list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

output-> result = ["a1", "b2", "c1", "d3", "a2"]

i.e. output format should be <value><count>

I have tried this solution, but its not giving expected output

def find_repetitive_elem(a):
    dict_obj = {}
    arr_obj = []
    for i in range(0, len(a)-1):
        count = 1    
        if a[i] == a[i+1]
            set_val = "%s%s" % (a[i], count+1)
        else:
            set_val = "%s%s" % (a[i], count)                
        arr_obj.append(set_val)

I'm trying to implement this using algorithm and not by python package's!

Upvotes: 2

Views: 95

Answers (6)

PradeepKumbhar
PradeepKumbhar

Reputation: 3421

This might look very basic, but this is how I understood the question:

Code:

list1 = ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

def find_repetitive_elem_continuously(arr):
    result = list()
    prev_ele = arr[0]
    count = 0
    for curr_ele in arr:
        if curr_ele != prev_ele:
            new_arr_ele = str(prev_ele)+""+str(count)
            result.append(new_arr_ele)
            prev_ele = curr_ele
            count = 1
        else:
            count += 1
    new_arr_ele = str(prev_ele)+""+str(count)
    result.append(new_arr_ele)
    print result

find_repetitive_elem_continuously(list1)

Output:

['a1', 'b2', 'c1', 'd3', 'a2']

Upvotes: 1

Alex
Alex

Reputation: 21766

You can use groupby from the itertools module to achieve this:

import itertools
list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]
["{}{}".format(key,sum( 1 for _ in group )) for key, group in itertools.groupby( list1 ) ]

As you have updated your answer stating that you do not want to use a Python package, this is the most elegant way I could think off that doesn't use external modules:

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

#output-> result = ["a1", "b2", "c1", "d3", "a2"]

def find_repetitive_elem(array):
    output=[]           
    for letter in array:
        existing_entry = {}
        new_entry = {}
        if len(output)>0:            
            existing_entry= output.pop();            
        if letter in existing_entry:            
            existing_entry[letter] +=1            
            output.append(existing_entry)
        else:            
            if existing_entry:            
                output.append(existing_entry)
            new_entry[letter]=1
            output.append(new_entry)                        

    return output

output = find_repetitive_elem(list1)        
print ["{}{}".format(key,value) for key,value in (entry.popitem() for entry in output)]

Upvotes: 2

sberry
sberry

Reputation: 132028

Not sure what you mean by

I'm trying to implement this using algorithm and not by python package's!

This is without using itertools (though that is a far better way to do this)

def compact(inval):
    last_char = None
    outval = []
    count = 0
    for l in inval + ['dummyvalue']:
        if last_char and last_char != l:
            outval.append("{}{}".format(last_char, count))
            count = 0
        last_char = l
        count += 1
    return outval

letters = ["a", "b", "b", "c", "d", "d", "d", "a", "a"]
print(compact(letters))

OUTPUT 

letters = ["a", "b", "b", "c", "d", "d", "d", "a", "a"] # yields
['a1', 'b2', 'c1', 'd3', 'a2']

letters = ["a", "b", "b", "c", "c", "d", "e", "d", "d", "d", "a", "a"] # yields
['a1', 'b2', 'c2', 'd1', 'e1', 'd3', 'a2']

Upvotes: 2

Noctis Skytower
Noctis Skytower

Reputation: 22011

Since you do not want to use any of Python's modules, here is a solution that does not use any (except for builtins implicitly). The code should be able to work on more than just strings.

def main():
    array = 'a', 'b', 'b', 'c', 'd', 'd', 'd', 'a', 'a'
    result = find_repetitive_elements(array)
    print(result)


def find_repetitive_elements(array):
    result, first, last, count = [], True, None, None
    for item in array:
        if first:
            first, last, count = False, item, 1
        elif item == last:
            count += 1
        else:
            result.append('{}{}'.format(last, count))
            last, count = item, 1
    if not first:
        result.append('{}{}'.format(last, count))
    return result


if __name__ == '__main__':
    main()

Upvotes: 2

Konstantin
Konstantin

Reputation: 25339

Here is my attempt without using itertools.groupby. The code relies on itertools.dropwhile iterator, but it is quite easy to replace it and I leave this as an exercise for you.

import itertools

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

result = []
lst = list1.copy()

while len(lst)>0:
    current_char = lst[0]
    length = len(lst)
    lst = list(itertools.dropwhile(lambda x: x==current_char, lst))
    result.append("{}{}".format(current_char, length-len(lst)))

print(result)

See code output at http://ideone.com/Cd1Y8B

Upvotes: 0

gtlambert
gtlambert

Reputation: 11961

You could use itertools.groupby:

from itertools import groupby

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

answer = [group[0] + str(len(list(group[1]))) for group in groupby(list1)]
print(answer)

Output

['a1', 'b2', 'c1', 'd3', 'a2']

Upvotes: 4

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